Crisp Fuzzy Implications

Abstract

Implications play an important role in fuzzy logics as they can be used both in practical and theoretical works. There exist many works in the literature where fuzzy implications behave in a crisp manner, i.e., implications that map to either zero or one. In this sense, we call those implications as crisp fuzzy implications and our goal is to study some their main features.

1 Introduction

A great deal of studies involving fuzzy implications can be found in the literature over the last years [1, 2, 4,5,6, 12]. Fuzzy implications are interesting from the theoretical point of view to its use on a variety of applications. For instance, they can be used to perform any fuzzy “if-then” rule in fuzzy systems and inference processes, which basically combine membership functions with the control rules to derive the fuzzy output. Regarding the theoretical aspect, many works have also been done aiming to generalize the traditional implication into fuzzy logic, explaining why there exists so many classes of fuzzy implications. The existence of those classes of fuzzy implications is justified by the fact that depending on the context or/and on the rules and their behavior, different implications with different properties can be adequate.

In the literature, it is possible to find examples of fuzzy implications with a crisp behavior, i.e., fuzzy implications that always map to either 0 or 1. For instance, in [13], it was defined two crisp-valued operators, named standard sharp and standard strict, as follows:

• 1. Standard sharp

  $$\begin{aligned} I_s(x, y) = {\left\{ \begin{array}{ll} 1, &{} \text { if } x < 1 \text { or } y = 1 \\ 0, &{} \text { otherwise } \end{array}\right. } \end{aligned}$$

• 2. Standard strict (also called Rescher-Gaines implication [8])

  $$\begin{aligned} I_G(x, y) = {\left\{ \begin{array}{ll} 1, &{} \text { if } x \le y \\ 0, &{} \text { otherwise } \end{array}\right. } \end{aligned}$$

Those implications were used in various applications, for instance, in the domains of approximate reasoning [8], relational databases [9], fuzzy control [10, 16], face recognition [17].

Thus, in this paper we intend to study the class of crisp fuzzy implications, i.e. fuzzy implications which always map to 0 or 1.

The paper is organized as follows. Section 2 summarizes some of the basic concepts demanded to understand the proposal in this work, including the concept of fuzzy implication and related properties. The study of crisp fuzzy implications is done in Sect. 3, including the most important results. At last, we finish in Sect. 4 with our final conclusions.

2 Preliminaries

Definition 1

A function $T: [0, 1]^{2} \rightarrow [0, 1]$ is said to be a triangular norm (t-norm, for short) if it satisfies the following conditions, for all $x, y, z \in [0, 1]$:

• (T1) Symmetry: $T(x,y) = T(y,x)$;

• (T2) Associativity: $T(x,T(y,z)) = T(T(x,y),z)$;

• (T3) Monotonicity: If $x_{1}\le x_{2}$ and $y_{1}\le y_{2}$ then $T(x_{1},y_{1}) \le T(x_{2},y_{2})$;

• (T4) 1-identity: $T(x,1) = x$.              (boundary condition)

In fuzzy logic, the conjunction is often represented by a t-norm. The standard fuzzy conjunction $T_M: [0, 1]^{2} \rightarrow [0, 1]$ given by $T_M(x, y) = min\{x,y\}$ is the only idempotent t-norm (see [11] - Theorem 3.9).

Proposition 1

[3]. Let T be a t-norm. Then $T(0,y) = 0$, for each $y \in [0, 1]$.

Definition 2

A t-norm T is called positive if satisfies the following condition: $T(x, y) = 0$ if and only if $x = 0$ or $y = 0$.

Definition 3

A function $S: [0, 1]^{2} \rightarrow [0, 1]$ is said to be a triangular conorm (t-conorm, for short) if it satisfies the following conditions, for all $x, y, z \in [0, 1]$:

• (S1) Symmetry: $S(x,y) = S(y,x)$;

• (S2) Associativity: $S(x,S(y,z)) = S(S(x,y),z)$;

• (S3) Monotonicity: If $x_{1}\le x_{2}$ and $y_{1}\le y_{2}$ then $S(x_{1},y_{1}) \le S(x_{2},y_{2})$;

• (S4) 0-identity: $S(x,0) = x$. (boundary condition)

From an axiomatic point of view, the difference between t-norms and t-conorms is just their boundary conditions.

Definition 4

A function $N: [0, 1] \rightarrow [0, 1]$ is called a fuzzy negation if

• (N1) N is antitonic, i.e. $N(x) \le N(y)$ whenever $y \le x$;

• (N2) $N(0) = 1$ and $N(1) = 0$.

  A fuzzy negation N is strict if

• (N3) N is continuous and

• (N4) $N(x) < N(y)$ whenever $y < x$.

  A fuzzy negation N is strong if

• (N5) $N(N(x)) = x$, for each $x \in [0, 1]$.

  A fuzzy negation N is crisp if

• (N6) $N(x) \in \{0,1\}$, for all $x \in [0, 1]$ (see [7]).

  A fuzzy negation N is frontier if it satisfies the following property:

• (N7) $N(x) \in \{0, 1\}$ if and only if $x = 0$ or $x = 1$.

Remark 1

By [7], a fuzzy negation $N: [0, 1] \rightarrow [0, 1]$ is crisp if and only if there exists $\alpha \in \left[ 0,1 \right)$ such that $N = N_\alpha$ or there exists $\alpha \in \left( 0,1 \right]$ such that $N = N^\alpha$, where

Theorem 1

[1]. If a function $N: [0, 1] \rightarrow [0, 1]$ satisfies (N1) and (N5), then it also satisfies (N2) and (N3). Moreover, N is a bijection, i.e., it satisfies (N4).

Corollary 1

[1]. Every strong negation is strict.

Definition 5

A function $I: [0, 1]^{2} \rightarrow [0, 1]$ is a fuzzy implication if the following properties are satisfied, for all $x, y, z \in [0, 1]$:

• (I1) If $x\le z$ then $I(x,y) \ge I(z,y)$;

• (I2) If $y\le z$ then $I(x,y) \le I(x,z)$;

• (I3) $I(0,y) = 1$;

• (I4) $I(x,1) = 1$;

• (I5) $I(1,0) = 0$.

The set of all fuzzy implications will be denoted by $\mathcal {FI}$.

Definition 6

Let $I \in \mathcal {FI}$. The function $N_{I}: [0, 1] \rightarrow [0, 1]$ defined by $N_{I}(x) = I(x,0)$, $x \in [0, 1]$ is called the natural negation of I or the negation induced by I.

Definition 7

Let T be a t-norm, S a t-conorm and N a fuzzy negation, then:

• A function $I: [0, 1]^{2} \rightarrow [0, 1]$ is called a (S, N)-implication (denoted by $I_{S,N}$) if $I(x,y) = S(N(x), y)$.

• A function $I: [0, 1]^{2} \rightarrow [0, 1]$ is called an R-implication (denoted by $I_T$) if $I(x,y) = sup\{t \in [0, 1] \mid T(x,t) \le y\}$.

• A function $I: [0, 1]^{2} \rightarrow [0, 1]$ is called a QL-implication (denoted by $I_{S,N,T}$) if $I(x,y) = S(N(x), T(x,y))$.

• A function $I: [0, 1]^{2} \rightarrow [0, 1]$ is called a D-implication (denoted by $I_{S,T,N}$) if $I(x,y) = S(T(N(x),N(y)), y)$.

Definition 8

A fuzzy implication I is said to satisfy:

• (i) the exchange principle if, for all $x, y, z \in [0, 1]$

  $I(x, I(y, z)) = I(y, I(x, z));$              (EP)

• (ii) the left neutrality property, if

  $I(1, y) = y, \ \ \ \ \ \ \ y \in [0, 1];$              (NP)

• (iii) the identity principle, if

  $I(x, x) = 1, \ \ \ \ \ \ \ x \in [0, 1];$              (IP)

• (iv) the left-ordering property if, for all $x, y \in [0, 1]$

  $I(x, y) = 1$ whenever $x \le y;$              (LOP)

• (v) the right-ordering property if for all $x, y \in [0, 1]$

  $I(x, y) \ne 1$ whenever $x > y$.              (ROP)

Definition 9

Let $I \in \mathcal {FI}$ and let N be a fuzzy negation. I is said to satisfy the:

• (i) contraposition law (or in other words, the contrapositive symmetry) with respect to N, if

  $I(x, y) = I(N(y), N(x)), \ x, y \in [0, 1];$              (CP)

• (ii) left contraposition law with respect to N, if

  $I(N(x), y) = I(N(y), x), \ x, y \in [0, 1];$              (L-CP)

• (iii) right contraposition law with respect to N, if

  $I(x, N(y)) = I(y, N(x)), \ x, y \in [0, 1]$.              (R-CP)

If I satisfies the (left, right) contrapositive symmetry with respect to a specific N, then we will denote this by $L{-}CP(N)$, $R{-}CP(N)$ and CP(N), respectively.

In [14, 15], Pinheiro et al. introduced a new class of implication, named (T, N)-implications which was defined by means of fuzzy negations and a t-norm.

Definition 10

[15]. Let N and $N'$ be fuzzy negations and T be a t-norm. The function $I_{T,N}^{N'}$ defined by $I_{T,N}^{N'}(x,y) = N'(T(x, N(y)))$ is called a $\mathbf {(N',T,N)}$-implication.

Actually, in [15] for (T, N)-implications we had $N'=N$ which is different from the previous definition where we have different negations. In order to avoid misunderstanding between definitions, from here forth, implications defined according to Definition 10 we be called $(N',T,N)$-implications.

3 Crisp Fuzzy Implications

In classical logic there is only one bivalent implication, however in the fuzzy setting the notion of bivalence gives rise to an uncountable family of such implications. They are called here Crisp fuzzy implication.

Definition 11

Let $I : [0, 1]^2 \rightarrow [0, 1]$ be a fuzzy implication. We say that I is a crisp fuzzy implication if $I(x,y) \in \{0,1\}$ for all $x, y \in [0, 1]$.

Proposition 2

Let $I : [0, 1]^2 \rightarrow [0, 1]$ be a fuzzy implication. Then I is crisp if and only if one of the following conditions are satisfied, for all $x, y \in [0, 1]$:

• (C1) If there exists $\alpha \in (0,1]$ and $\beta \in [0, 1)$ such that $I(x,y) = I_{\alpha , \beta } (x, y)$, where

  $$\begin{aligned} I_{\alpha , \beta } (x, y) = {\left\{ \begin{array}{ll} 0, &{} { if }\ x \ge \alpha \ { and }\ y \le \beta \\ 1, &{} { otherwise } \end{array}\right. }; \end{aligned}$$

• (C2) If there exists $\alpha \in [0,1)$ and $\beta \in (0, 1]$ such that $I(x,y) = I^{\alpha , \beta } (x, y)$, where

  $$\begin{aligned} I^{\alpha , \beta } (x, y) = {\left\{ \begin{array}{ll} 0, &{}\ { if }\ x > \alpha \ { and }\ y < \beta \\ 1, &{} { otherwise } \end{array}\right. }; \end{aligned}$$

• (C3) If there exists $\alpha , \beta \in (0,1]$ such that $I(x,y) = {I_{\alpha }}^{\beta } (x, y)$, where

  $$\begin{aligned} {I_{\alpha }}^{\beta } (x, y) = {\left\{ \begin{array}{ll} 0, &{} { if }\ x \ge \alpha \ { and }\ y < \beta \\ 1, &{} { otherwise } \end{array}\right. }; \end{aligned}$$

• (C4) If there exists $\alpha , \beta \in [0,1)$ such that $I(x,y) = {I^{\alpha }}_{\beta } (x, y)$, where

  $$\begin{aligned} {I^{\alpha }}_{\beta } (x, y) = {\left\{ \begin{array}{ll} 0, &{} { if }\ x > \alpha \ \text { and }\ y \le \beta \\ 1, &{} { otherwise } \end{array}\right. }. \end{aligned}$$

Proof

First, suppose I is crisp, then as $I(0,0) = 1$ and $I(1,0) = 0$, we have by (I1) that there exists: (1) $\alpha \in (0,1]$ such that $I (x, 0) = {\left\{ \begin{array}{ll} 0, &{} \text { if } x \ge \alpha \\ 1, &{} \text { otherwise } \end{array}\right. }$ or (2) $\alpha \in [0,1)$ such that $I (x, 0) = {\left\{ \begin{array}{ll} 0, &{} \text { if } x > \alpha \\ 1, &{} \text { otherwise } \end{array}\right. }.$ By (I2), we have for case (1) that there exist: $(i)_1$ $\beta \in [0, 1)$ such that $I(x,y) = {\left\{ \begin{array}{ll} 0, &{} \text { if } x \ge \alpha \text { and } y \le \beta \\ 1, &{} \text { otherwise } \end{array}\right. }$ or $(ii)_1$ $\beta \in (0, 1]$ such that $I(x,y) = {\left\{ \begin{array}{ll} 0, &{} \text { if } x \ge \alpha \text { and } y < \beta \\ 1, &{} \text { otherwise } \end{array}\right. }.$ Hence, $I(x,y) = I_{\alpha , \beta } (x, y)$ or $I(x,y) = {I_{\alpha }}^{\beta } (x, y)$, for all $x, y \in [0, 1]$, respectively. Similarly, still by (I2), for case (2) there exist: $(i)_2$ $\beta \in [0, 1)$ such that $I(x,y) = {\left\{ \begin{array}{ll} 0, &{} \text { if } x > \alpha \text { and } y \le \beta \\ 1, &{} \text { otherwise } \end{array}\right. }$ or $(ii)_2$ $\beta \in (0, 1]$ such that $I(x,y) = {\left\{ \begin{array}{ll} 0, &{} \text { if } x > \alpha \text { and } y < \beta \\ 1, &{} \text { otherwise } \end{array}\right. }.$ Hence, $I(x,y) = {I^{\alpha }}_{\beta } (x, y)$ or $I(x,y) = I^{\alpha ,\beta } (x, y)$, for all $x, y \in [0, 1]$, respectively.

The reciprocal case follows straightforward.

Definition 12

Let I be a crisp fuzzy implication. Independently from I being of type C1, C2, C3 or C4, the pair $(\alpha , \beta )$ is called the threshold pair of I.

In the following proposition, we can observe that we can obtain a crisp fuzzy implications from any fuzzy implication I.

Proposition 3

Let $I \in \mathcal {FI}$. Then, for any $\gamma \in \left( 0,1\right]$, $I_{\gamma }(x, y) = {\left\{ \begin{array}{ll} 1, &{} { if }\ I(x, y) \ge \gamma \\ 0, &{} { if }\ I(x, y) < \gamma \end{array}\right. }$ is a crisp fuzzy implication.

Proof

We will first prove that $I_{\gamma }$ satisfies the conditions demanded in Definition 5. Indeed,

• ($I_{\gamma }$1) For all $x, y, z \in [0, 1]$, such that $x \le y$, we have by (I1) that $I(y, z) \le I(x, z)$. We will analyze the following cases: (1) If $I(x, z) \ge I(y, z) \ge \gamma$, then $I_{\gamma }(x, z) = I_{\gamma }(y, z) = 1$; (2) If $I(x, z) \ge \gamma > I(y, z)$, then $I_{\gamma }(x, z) = 1 > 0 = I_{\gamma }(y, z)$ and (3) if $\gamma > I(x, z) \ge I(y, z)$, then $I_{\gamma }(x, z) = I_{\gamma }(y, z)$ = 0. Therefore, $I_{\gamma }$ satisfies (I1).

• ($I_{\gamma }$2) For all $x, y, z \in [0, 1]$, such that $y \le z$, we have by (I2) that $I(x, y) \le I(x, z)$. We will analyze the following cases: (1) If $\gamma \le I(x, y) \le I(x, z)$, then $I_{\gamma }(x, y) = I_{\gamma }(x, z) = 1$; (2) If $I(x, y) < \gamma \le I(x, z)$, then $I_{\gamma }(x, y) = 0 < 1 = I_{\gamma }(x, z)$ and (3) if $I(x, y) \le I(x, z) < \gamma$, then $I_{\gamma }(x, y) = I_{\gamma }(x, z)$ = 0. Therefore, $I_{\gamma }$ satisfies (I2).

• ($I_{\gamma }$3) For all $y \in [0, 1]$, we have by (I3) that $I(0, y) =1$. So, $I(0, y) \ge \gamma$ and thereby $I_{\gamma }(0, y) = 1$. Therefore, $I_{\gamma }$ satisfies (I3).

• ($I_{\gamma }$4) For all $x \in [0, 1]$, we have by (I4) that $I(x, 1) =1$. So, $I(x, 1) \ge \gamma$ and thereby $I_{\gamma }(x, 1) = 1$. Therefore, $I_{\gamma }$ satisfies (I4).

• ($I_{\gamma }$5) By (I5), $I(1, 0) = 0$. So, $I(1, 0) < \gamma$ and thereby $I_{\gamma }(1, 0) = 0$. Therefore, $I_{\gamma }$ satisfies (I5).

We conclude that $I_{\gamma }$ is a fuzzy implication. As, $I_{\gamma }(x,y) \in \{0,1\}$ for all $x, y \in [0, 1]$, then $I_{\gamma }$ is a crisp fuzzy implication.

Notice that if we take $\gamma \in \left[ 0,1\right)$, and $I^{\gamma }(x, y) = {\left\{ \begin{array}{ll} 1, &{} \text { if } I(x, y) > \gamma \\ 0, &{} \text { if } I(x, y) \le \gamma \end{array}\right. }$, $I^{\gamma }$ is also a crisp fuzzy implication.

Proposition 4

Let $I_{T,N}^{N'}$ be a $(N',T,N)$-implication. Then $I_{T,N}^{N'}$ is crisp if and only if $N'$ is a crisp fuzzy negation.

Proof

Suppose $N'$ is not crisp, then there is $z \in (0,1)$ such that $N'(z) \notin \{0,1\}$. So, for any t-norm T and any fuzzy negation N we have that $T(z, N(0)) = T(z,1) = z$, thus $I_{T,N}^{N'}(z,0) = N'(T(z, N(0))) = N'(z) \notin \{0,1\}$. Therefore $I_{T,N}^{N'}$ is not crisp. Conversely, if $N'$ is crisp then, for any t-norm T and any fuzzy negation N, $I_{T,N}^{N'}(x,y) = N'(T(x, N(y))) \in \{0,1\}$.

Corollary 2

Let $I_{T}^{N}$ be a (T, N)-implication. Then $I_{T}^{N}$ is crisp if and only if N is a crisp fuzzy negation.

Proposition 5

Let I be a (T, N)-implication for any crisp negation N and any t-norm T. Then:

• (i) I satisfies (LOP);

• (ii) I does not satisfy (ROP).

Proof

Since N is crisp we have, by Corollary 2, that $I=I_{T}^{N}$ is crisp. Note that:

$$\begin{aligned} I(x,y)=N(T(x,N(y)))={\left\{ \begin{array}{ll} 1, &{} \text { if } N(y)=0 \\ N(x), &{} \text { if } N(y)=1 \end{array}\right. }. \end{aligned}$$

Then,

• (i) For all $x,y \in [0, 1]$ such that $x \le y$, since N is crisp, we will analyze two cases:

  (1) if there exists $\alpha \in \left[ 0,1 \right)$ such that $N = N_\alpha$, then

  $$\begin{aligned} I(x,y)={\left\{ \begin{array}{ll} 1, &{} \text { if } y > \alpha \\ N(x), &{} \text { if } y\le \alpha \end{array}\right. }. \end{aligned}$$

  (1)

  For $y\le \alpha$, as $x \le y$, then $x \le \alpha$. So $N_\alpha (x) = 1$ and hence $I(x,y)=1$.

  (2) If there exists $\alpha \in \left( 0,1 \right]$ such that $N = N^\alpha$, then

  $$\begin{aligned} I(x,y)={\left\{ \begin{array}{ll} 1, &{} \text { if } y \ge \alpha \\ N(x), &{} \text { if } y < \alpha \end{array}\right. }. \end{aligned}$$

  (2)

  For $y < \alpha$, as $x \le y$, then $x < \alpha$. So $N_\alpha (x) = 1$ and therefore, $I(x,y)=1$. Thus, I satisfies (LOP).

• (ii) We will analyze two cases again:

  (1) if $N = N_\alpha$ for some $\alpha \in \left[ 0,1 \right)$, then there exists $x,y \in [0, 1]$ such that $x> y > \alpha$. So, by Eq. (1), $I(x,y) = 1$.

  (2) If $N = N^\alpha$ for some $\alpha \in \left( 0,1 \right]$, then there exists $x,y \in [0, 1]$ such that $y< x < \alpha$. So, by Eq. (2) $I(x,y) = N^\alpha (x) = 1$. In any case, there exists $x > y$, but $I(x,y) = 1$, therefore I does not satisfy (ROP).

Proposition 6

Let I be a crisp fuzzy implication. Then:

• (i) I satisfies (EP);

• (ii) I satisfies $R{-}CP(N_{I})$, where $N_{I}$ is the natural negation of I;

• (iii) I does not satisfy (NP);

Proof

• (i) Since I is crisp, then by Proposition 2, I satisfies one of the conditions (C1), (C2), (C3) or (C4). If I satisfies (C1), then there exists $\alpha \in (0,1]$ and $\beta \in [0, 1)$ such that $I(x,y) = I_{\alpha , \beta } (x, y)$. So, (1) for $z \le \beta$, we have $I(y, z) = {\left\{ \begin{array}{ll} 0, &{} \text { if } y \ge \alpha \\ 1, &{} \text { if } y < \alpha \end{array}\right. }.$ Thus,

  $$\begin{aligned} I(x, I(y, z))= & {} {\left\{ \begin{array}{ll} I(x, 0), &{} \text { if } y \ge \alpha \\ I(x, 1), &{} \text { if } y < \alpha \end{array}\right. } = {\left\{ \begin{array}{ll} 0, &{} \text { if } x \ge \alpha \text { and } y \ge \alpha \\ 1, &{} \text { otherwise } \end{array}\right. }, \end{aligned}$$

  and we also have $I(x, z) = {\left\{ \begin{array}{ll} 0, &{} \text { if } x \ge \alpha \\ 1, &{} \text { if } x < \alpha \end{array}\right. }.$ Thus,

  $$\begin{aligned} I(y, I(x, z))= & {} {\left\{ \begin{array}{ll} I(y, 0), &{} \text { if } x \ge \alpha \\ I(y, 1), &{} \text { if } x < \alpha \end{array}\right. } = {\left\{ \begin{array}{ll} 0, &{} \text { if } x \ge \alpha \text { and } y \ge \alpha \\ 1, &{} \text { otherwise } \end{array}\right. }. \end{aligned}$$

  Then, for $z \le \beta$, (EP) is satisfied. Now, (2) for $z > \beta$, we have $I(y, z) = I(x, z) = 1$. So $I(x, I(y, z)) = I(x, 1) = 1 = I(y, 1) = I(y, I(x, z))$.

  Cases (C2), (C3) and (C4) are similar to the previous one.Therefore, I satisfy (EP).

• (ii) Indeed, for all $x, y \in [0, 1]$,

  $I(x, N_{I}(y)) = I(x, I(y, 0)) {\mathop {=}\limits ^{(EP)}} I(y, I(x, 0)) = I(y, N_{I}(x))$.

• (iii) As $I(x,y) \in \{0,1\}$, since I is crisp then, for all $y \in (0,1)$, $I(x,y) \ne y$.

Proposition 7

Let I be a crisp fuzzy implication with $(\alpha ,\beta )$ as its threshold. If $\beta < \alpha$ then I satisfies (IP).

Proof

Indeed, if I is of type (C1) then there is no $x \in [0, 1]$ such that $x \ge \alpha$ and $x \le \beta$ simultaneously, since $\beta < \alpha$. So $I(x, x) = 1$, for all $x \in [0, 1]$. Similarly, if I is of type (C2), (C3) or (C4) we prove that $I(x,x)=1$. Therefore, in any case, I satisfies (IP).

Proposition 8

Let I be a crisp fuzzy implication with $(\alpha ,\beta )$ as its threshold. If $\alpha < \beta$ then I does not satisfy (IP).

Proof

Indeed, because there exists x between $\alpha$ and $\beta$ such that $I(x,x) = 0$, therefore I does not satisfy (IP).

Proposition 9

Let I be a crisp fuzzy implication with $(\alpha ,\beta )$ as its threshold. If $\alpha = \beta$ then:

• (i) If I is of type (C1) then I does not satisfy (IP);

• (ii) If I is of type (C2), (C3) or (C4) then I satisfies (IP).

Proof

In fact,

• (i) There is $x = \alpha$ such that $I(x,x)=0 \ne 1$, so I does not satisfy (IP).

• (ii) If I is of type (C2) then there is no $x \in [0, 1]$ such that $x > \alpha$ and $x < \beta$ simultaneously, since $\beta = \alpha$. So $I(x, x) = 1$, for all $x \in [0, 1]$. Similarly, if I is of type (C3) or (C4) we prove that $I(x,x)=1$. Therefore, in any case, I satisfies (IP).

In [4], the conditions under which the Boolean-like law holds for some classes of fuzzy implications were given. Here, we prove that it is valid for a crisp fuzzy implication whenever (IP) is satisfied.

Proposition 10

Let I be a crisp fuzzy implication. Then:

• (i) If I satisfies (IP) then $I(x, I(y,x)) = 1$, for all $x, y \in [0, 1]$;

• (ii) If I does not satisfy (IP) then there are $x, y \in [0, 1]$ such that $I(x, I(y,x)) \ne 1$.

Proof

Indeed,

• (i) If I satisfies (IP) then $I(x, x) = 1$, for all $x \in [0, 1]$, so by Proposition 6: $I(x, I(y, x)) {\mathop {=}\limits ^{(EP)}} I(y, I(x, x)) = I(y, 1) = 1.$

• (ii) If I does not satisfy (IP) then there exist $x \in (0,1)$ such that $I(x,x) = 0$. So, for $y = 1$, $I(x, I(1,x)) = I(1, I(x,x)) = I(1, 0) = 0 \ne 1$.

Corollary 3

Let I be a crisp fuzzy implication with $(\alpha ,\beta )$ as its threshold.

• (i) If $\beta < \alpha$ then $I(x, I(y,x)) = 1$, for all $x, y \in [0, 1]$;

• (ii) If $\alpha < \beta$ then there exists $x, y \in [0, 1]$ such that $I(x, I(y,x)) \ne 1$;

• (iii) If $\alpha = \beta$ then there exists $x, y \in [0, 1]$ such that $I(x, I(y,x)) \ne 1$ whenever I is of type (C1). And $I(x, I(y,x)) = 1$, for all $x, y \in [0, 1]$ whenever I is of type (C2), (C3) or (C4).

Proof

It follows straight from Propositions 7, 8 and 9.

Definition 13

Let I be a crisp fuzzy implication with $(\alpha ,\beta )$ as its threshold and let N be a fuzzy negation. We say that IN is a dual crisp fuzzy implication of I with respect to N , or dual NCrisp, if it satisfies one of the following types, for all $x, y \in [0, 1]$:

• (NC1) $IN(x, y) =\underline{} I_{N(\beta ), N(\alpha )} (x, y)$, whenever I satisfies (C1);

• (NC2) $IN(x, y) = I^{N(\beta ), N(\alpha )} (x, y)$, whenever I satisfies (C2);

• (NC3) $IN(x, y) = {I^{N(\beta )}}_{N(\alpha )}(x, y)$, whenever I satisfies (C3);

• (NC4) $IN(x, y) = {I_{N(\beta )}}^{N(\alpha )}(x, y)$, whenever I satisfies (C4).

Definition 14

Let I be a crisp fuzzy implication and N be a fuzzy negation. I is said to be a

• (i) Crisp-CP with respect to N, if

  $I(x, y) = IN(N(y), N(x)),$              (C-CP)

• (ii) Crisp Left-CP with respect to N, if

  $I(N(x), y) = IN(N(y), x),$              (C-LCP)

• (iii) Crisp Right-CP with respect to N, if

  $I(x, N(y)) = IN(y, N(x)),$              (C-RCP)

where IN is its dual NCrisp.

Proposition 11

Let I be a crisp fuzzy implication and N be a fuzzy negation. If N is strict, then I is C-CP with respect to N.

Proof

• (1) If I satisfies (C1), then, by Proposition 2, there exist $\alpha \in (0,1]$ and $\beta \in [0, 1)$ such that $I(x,y) = I_{\alpha , \beta } (x, y)$. So, as N is strict, $x \ge \alpha$ if and only if $N(x) \le N(\alpha )$ and $y \le \beta$ if and only if $N(y) \ge N(\beta )$. Therefore, by (NC1)

  $$\begin{aligned}&IN (N(y), N(x)) = I_{N(\beta ),N(\alpha ) } (N(y), N(x))\\&= {\left\{ \begin{array}{ll} 0, &{} \text { if } N(y) \ge N(\beta ) \text { and } N(x) \le N(\alpha ) \\ 1, &{} \text { otherwise } \end{array}\right. }\\&= {\left\{ \begin{array}{ll} 0, &{} \text { if } x \ge \alpha \text { and } y \le \beta \\ 1, &{} \text { otherwise } \end{array}\right. } = I_{\alpha , \beta } (x, y) = I(x,y). \end{aligned}$$

• (2) If I satisfies (C2), then, by Proposition 2, there exist $\alpha \in [0,1)$ and $\beta \in (0, 1]$ such that $I(x,y) = I^{\alpha , \beta } (x, y)$. So, as N is strict, $x > \alpha$ if and only if $N(x) < N(\alpha )$ and $y < \beta$ if and only if $N(y) > N(\beta )$. Therefore, by (NC2)

  $$\begin{aligned}&IN (N(y), N(x)) = I^{N(\beta ),N(\alpha ) } (N(y), N(x))\\&= {\left\{ \begin{array}{ll} 0, &{} \text { if } N(y)> N(\beta ) \text { and } N(x)< N(\alpha ) \\ 1, &{} \text { otherwise } \end{array}\right. }\\&= {\left\{ \begin{array}{ll} 0, &{} \text { if } x > \alpha \text { and } y < \beta \\ 1, &{} \text { otherwise } \end{array}\right. } = I^{\alpha , \beta } (x, y) = I(x,y). \end{aligned}$$

• (3) If I satisfies (C3), then, by Proposition 2, there exist $\alpha , \beta \in (0,1]$ such that $I(x,y) = {I_{\alpha }}^{\beta } (x, y)$. So, as N is strict, $x \ge \alpha$ if and only if $N(x) \le N(\alpha )$ and $y < \beta$ if and only if $N(y) > N(\beta )$. Therefore, by (NC3)

  $$\begin{aligned}&IN (N(y), N(x)) = {I^{N(\beta )}}_{N(\alpha )} (N(y), N(x))\\&= {\left\{ \begin{array}{ll} 0, &{} \text { if } N(y) > N(\beta ) \text { and } N(x) \le N(\alpha ) \\ 1, &{} \text { otherwise } \end{array}\right. }\\&= {\left\{ \begin{array}{ll} 0, &{} \text { if } x \ge \alpha \text { and } y < \beta \\ 1, &{} \text { otherwise } \end{array}\right. } = {I_{\alpha }}^{\beta } (x, y) = I(x,y). \end{aligned}$$

• (4) If I satisfies (C4), then, by Proposition 2, there exist $\alpha , \beta \in [0,1)$ such that $I(x,y) = {I^{\alpha }}_{\beta } (x, y)$. So as N is strict, $x > \alpha$ if and only if $N(x) < N(\alpha )$ and $y \le \beta$ if and only if $N(y) \ge N(\beta )$. Therefore, by (NC4)

  $$\begin{aligned}&IN (N(y), N(x)) = {I_{N(\beta )}}^{N(\alpha )} (N(y), N(x))\\&= {\left\{ \begin{array}{ll} 0, &{} \text { if } N(y) \ge N(\beta ) \text { and } N(x) < N(\alpha ) \\ 1, &{} \text { otherwise } \end{array}\right. }\\&= {\left\{ \begin{array}{ll} 0, &{} \text { if } x > \alpha \text { and } y \le \beta \\ 1, &{} \text { otherwise } \end{array}\right. } = {I^{\alpha }}_{\beta } (x, y) = I(x,y). \end{aligned}$$

Therefore, in any case, I is C-CP.

Notice that for C-LCP and C-RCP, the requirements are different from Proposition 11. The proof for both is analogous as we can see in the following proposition.

Proposition 12

Let I be a crisp fuzzy implication and N be a fuzzy negation. If N is strong, then I is C-LCP and C-RCP with respect to N.

Proof

Straightforward.

Finally, we also studied that is impossible for some implications to be crisp fuzzy implications.

Proposition 13

None of the following classes of fuzzy implications (S, N)-, R-, QL- and D-implication is a crisp fuzzy implication.

Proof

In fact,

• (1) if I is an (S, N)-implication, then there exist a t-conorm S and a fuzzy negation N such that $I(x,y) = S(N(x), y)$ for all $x, y \in [0, 1]$. In particular, for $x = 1$, $I(1, y) = S( N(1), y) = S(0,y) {\mathop {=}\limits ^{(S4)}} y$. So, for all $y \in (0,1)$, $I(1, y) \notin \{0,1\}$. Therefore, I is not crisp.

• (2) If I is an R-implication, then there exists a t-norm T such that $I(x,y) = sup\{t \in [0, 1] \mid T(x,t) \le y \}$ for all $x, y \in [0, 1]$. In particular, for $x = 1$, $I(1, y) = sup\{t \in [0, 1] \mid T(1,t) \le y \} {\mathop {=}\limits ^{(T4)}} sup\{t \in [0, 1] \mid t \le y \} = y.$ So, for all $y \in (0,1)$, $I(1, y) \notin \{0,1\}$. Therefore, I is not crisp.

• (3) If I is a QL-implication, then there exist a t-norm T, a t-conorm S and a fuzzy negation N such that $I(x,y) = S(N(x), T(x,y))$ for all $x, y \in [0, 1]$. In particular, for $x = 1$, $I(1, y) = S(N(1), T(1,y)) = S(0, T(1,y)) {\mathop {=}\limits ^{(S4)}} T(1,y) {\mathop {=}\limits ^{(T4)}} y$. So, for all $y \in (0,1)$, $I(1, y) \notin \{0,1\}$. Therefore, I is not crisp.

• (4) If I is a D-implication, then there exist a t-norm T, a t-conorm S and a fuzzy negation N such that $I(x,y) = S(T(N(x), N(y)), y)$ for all $x, y \in [0, 1]$. In particular, for $x = 1$, $I(1, y) = S(T(N(1), N(y)), y) = S(T(0, N(y)), y) = S(0,y) {\mathop {=}\limits ^{(S4)}} y.$ So, for all $y \in (0,1)$, $I(1, y) \notin \{0,1\}$. Therefore, I is not crisp.

4 Final Remarks

One can find many examples of studies which use fuzzy implications with crisp behavior such as [8,9,10, 17]. Our purpose in this work was to study those fuzzy implications which always map to 0 or 1, therefore named crisp fuzzy implications. We provided a characterization for those implications, presenting four possible classes (Proposition 2) and studied some properties and conditions under which fuzzy implications need in order to be considered crisp.